<p>80. The range of \(a\) for which the equation \(x^2 + ax - 4 = 0\) has its smaller root in the interval \((-1, 2)\) is</p>
<p>(1) \((-\infty, -3)\)</p>
<p>(2) \((0, 3)\)</p>
<p>(3) \((0, \infty)\)</p>
<p>(4) \((-\infty, -3) \cup (0, \infty)\)</p>
Step-by-Step Solution
Key Concept: For the smaller root to lie in (-1, 2), we need f(-1) and f(2) to have opposite signs (so a root exists between them), and the vertex of the parabola must be positioned correctly. Since the parabola opens upward, f(-1) > 0 and f(2) > 0 ensures the smaller root is in (-1, 2).
<p><strong>Step 1:</strong> For a quadratic f(x) = x² + ax - 4 with roots r₁ < r₂, we need the smaller root r₁ ∈ (-1, 2).</p><p><strong>Step 2:</strong> Since the parabola opens upward and we want r₁ ∈ (-1, 2), we require f(-1) > 0 and f(2) > 0 (the function is positive at both boundaries, meaning the smaller root lies between them).</p><p><strong>Step 3:</strong> Calculate f(-1): f(-1) = (-1)² + a(-1) - 4 = 1 - a - 4 = -3 - a. For f(-1) > 0: -3 - a > 0 ⟹ a < -3</p><p><strong>Step 4:</strong> Calculate f(2): f(2) = (2)² + a(2) - 4 = 4 + 2a - 4 = 2a. For f(2) > 0: 2a > 0 ⟹ a > 0</p><p><strong>Step 5:</strong> We need a < -3 AND a > 0, which appears contradictory. Re-examine: For the smaller root in (-1, 2) with upward parabola, we need f(-1) > 0 and f(2) > 0 simultaneously is impossible. Instead, the correct condition is: f(-1) > 0 and f(2) > 0 with the vertex in (-1, 2), OR one value negative. The vertex is at x = -a/2. For r₁ ∈ (-1, 2): we need -1 < -a/2 < 2, giving -4 < a < 2.</p><p><strong>Step 6:</strong> Combined with discriminant Δ = a² + 16 > 0 (always true) and f(-1)·f(2) < 0 OR both positive with vertex in interval: The answer is <strong>a ∈ (-4, 0) ∪ (0, 2) = (-4, 2) \ {0}</strong> or simplified as <strong>-4 < a < 2</strong>.</p><p>∴ Answer: <strong>a ∈ (-4, 2)</strong></p>
Correct Answer: 4